1. ## Question! Binomial Series

Hi guys I can't do the last part of this question so thanks if anybody could help!

Find the first four terms, in ascending order, in the expansion of (1-2x)^-2. State the values of x for which this expansion is valid.
Hence,

(i) by choosing a suitable value of x, find sum to infinity (starting value r=1) of r/2^(r-1),
(i) show that [sq root 2 cos (2x+(pi/4))] / (1-2tanx)^2 is estimated to be 1+2x+2x^2, given that x is small enough for its third and higher powers to be ignored.

Many apologies for the 'messiness'!

2. Let $\displaystyle r$ be any real number and $\displaystyle |y|<1$, theny

$\displaystyle (1+y)^r = 1 + \frac{r}{1!}y+\frac{r(r-1)}{2!}y^2+\frac{r(r-1)(r-2)}{3!}y^3+...$

Now, we want to expand $\displaystyle (1-2x)^{-2}$ thus $\displaystyle r=-2$ and $\displaystyle y=-2x$. Hence we require that $\displaystyle |-2x|=|2x|<1$ that is, $\displaystyle |x|<\frac{1}{2}$.

Thus the binomial series look like:
$\displaystyle (1-2x)^{-2} = 1 - \frac{-2}{1!}(2y) + \frac{-2(-3)}{2!}(2y)^2 - \frac{-2(-3)(-4)}{3!} (2y)^3+$